Standard deviation and data comparison

Deviations measure distance from the mean

For values 2, 4, 4, 6, the mean is 4. Deviations are −2, 0, 0, 2. Their sum is zero, so averaging signed deviations would not measure spread. Square them instead: 4, 0, 0, 4. Their mean is 2, the variance under the whole-data-set convention. The standard deviation is square root of 2 ≈ 1.414.

For this convention,

σ = square root of ((∑(x-x bar)²)/n) = square root of ((∑ x²)/n-x bar²).

Use the divisor n for the standard deviation of the listed data set here. A sample-based estimator with divisor n-1 is a different statistic, not an interchangeable calculator display. Check which mode or output a task requests.

Frequencies count repeated values

With a frequency table, use n = ∑ f, mean ∑ fx/n and variance ∑ fx²/n-x bar². A class midpoint can replace x for grouped data, but then both the mean and standard deviation are estimates.

For midpoints 5, 15, 25, 35 and frequencies 4, 8, 6, 2, n = 20, estimated mean is 18 and ∑ fx² = 8100. Estimated variance is 8100/20-18² = 81, so estimated standard deviation is 9 minutes. Keep extra digits during the calculation to avoid a negative variance from premature rounding.

Changes of scale have predictable effects

Adding 10 to every value adds 10 to the mean but leaves every deviation unchanged, so standard deviation stays the same. Multiplying every value by 3 multiplies the mean by 3 and standard deviation by 3; variance is multiplied by 9.

A smaller standard deviation indicates less spread around the mean for comparable data in the same unit. It does not by itself establish a better outcome or prove that every value is closer to a target.

Check your understanding 1 Separate location from spread

A data set has mean 12 and standard deviation 2. Each value is transformed by y = 5x-3. Find the new mean and standard deviation.

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The mean becomes 5(12)-3 = 57. Standard deviation becomes 5(2) = 10. Subtracting 3 shifts all values together and does not change spread.

Report a comparison with its context

Two machines have the same mean output but standard deviations 1.2 and 3.8 units. The first is more consistent by this measure. Whether either meets a required tolerance needs the tolerance and the distribution; the standard deviations alone do not certify every item.

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